Question

Solve : $\int \frac{1}{1+\tan x}dx$

• A. $\frac{1}{2}[x+|(\cos x-\sin x\left)|\right]+C$
• B. $\frac{1}{2}[x+\ln |(\cos x+\sin x\left)|\right]+C$
• C. $\frac{1}{3}[x+\ln |(\cos x-\sin x\left)|\right]+C$
• D. $\frac{1}{4}[x+\ln |(\cos x-\sin x\left)|\right]+C$

Answer 1

The correct option is B $\frac{1}{2}[x+\ln |(\cos x+\sin x\left)|\right]+C$

$\int \frac{1}{1+\tan x}dx$

$\int \frac{1}{1+\frac{\sin x}{\cos x}}dx$

$\int \frac{\cos x}{\cos x+\sin x}dx$

$\frac{1}{2}\int (\frac{2\cos x}{\cos x+\sin x}-1+1)dx$

$\frac{1}{2}\int \left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)dx+\frac{1}{2}\int 1dx$

substitute $u=\cos x+\sin x\to du=(\cos x-\sin x)dx$

$\frac{1}{2}\frac{du}{u}+\frac{1}{2}x$

$\frac{1}{2}(\ln u+x)$

$\frac{1}{2}[x+\ln (\cos x+\sin x\left)\right]+C$